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Let $m,n$ be coprime square-free integers, $m\equiv 3(\mathrm{mod}4)$ and $n\equiv 2(\mathrm{mod}4)$. For the biquadratic extension $L=\mathbb{Q}(\alpha ,\beta )$, with ${\alpha }^2=m$ and ${\beta }^2=n$, an integral basis is $1,\alpha ,\beta ,(\beta +\alpha \beta )/2$. To prove spanning, write an algebraic integer as $(a+b\alpha +c\beta +d\alpha \beta )/2$. The field traces to the three quadratic subfields and the ring of integers of a quadratic field formula force $a,b,c,d\in \mathbb{Z}$. The field norm to $\mathbb{Q}(\alpha )$ has coefficients $(a^2+m{b}^2-n{c}^2-mn{d}^2)/4$ and $(ab-ncd)/2$. Their integrality gives $ab\equiv 0(\mathrm{mod}2)$ and $a^2-b^2-2(c^2+d^2)\equiv 0(\mathrm{mod}4)$. Thus $a,b$ are even and $c\equiv d(\mathrm{mod}2)$, exactly the asserted integer span. Conversely, the half-sum $w$ has $w^2=n(1+m)/4+(n/2)\alpha \in \mathbb{Z}[\alpha ]$, so integral dependence is transitive proves it is an algebraic integer. Rational linear independence is immediate from its nonzero $\alpha \beta$ coefficient.

Again put $\alpha =\sqrt{m_1}$ and $\beta =\sqrt{m_2}$. The ring of integers of each of the three quadratic subfields has integral basis $1,\alpha$, $1,\beta$, or $1,\alpha \beta$, respectively: their square-free radicands are respectively congruent to $3,2,2$ modulo four. The field traces of the algebraic integer $\gamma$ to those subfields areThey are algebraic integers, so comparison with each integral basis givesThis use of all three subfields is needed to control $d$ as well as the other coefficients.

The field norm to $\mathbb{Q}(\alpha )$ isIts two coefficients must be integers. The second condition says $ab$ is even, since $m_2$ is even. Reducing the numerator of the first coefficient modulo four givesReduction modulo two first gives $a\equiv b(\mathrm{mod}2)$. Since $ab$ is even, both $a$ and $b$ are even. The same congruence now gives $c^2+d^2\equiv 0(\mathrm{mod}2)$, or $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer has the formwith four integer coefficients.

For the converse, $1,\alpha ,\beta$ are algebraic integers, andHere $4\mid 1+m_1$ and $2\mid m_2$, so both coefficients are integers. The monic polynomial $X^2-w^2$ and integral dependence is transitive show that $w$ is an algebraic integer. The four displayed generators are rationally linearly independent, since their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/2$. Hence the integral basis for coprime biquadratic radicands congruent to three and two isIn this formula $\sqrt{m_1{m}_2}$ denotes the product $\alpha \beta$ of the chosen square roots; this convention also handles negative radicands.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$ and $F=\mathbb{Q}(\alpha )$. The hypotheses make $L/\mathbb{Q}$ a biquadratic extension. To check the degree of a field extension is four, if $\beta =x+y\alpha$ with rational $x,y$, squaring gives $xy=0$. The case $y=0$ would make $m_2$ a rational square; the case $x=0$ would make $m_2/m_1$ a rational square. Neither can occur for the specified distinct coprime square-free integers. Hence $[L:F]=2$.

The nonidentity field automorphism of $L/F$ fixes $\alpha$ and sends $\beta$ to $-\beta$. Thus, for $\gamma =(r+s\alpha )+(t+u\alpha )\beta$, the field norm and field trace areIf $\gamma$ is an algebraic integer, so is its conjugate, and their sum and product are algebraic integers. Conversely, put $T={\mathrm{Tr}}_{L/F}(\gamma )$ and $N=N_{L/F}(\gamma )$. The identityshows that $\gamma$ is integral over $\mathbb{Z}[T,N]$. If $T,N$ are algebraic integers, the ring generated by algebraic integers $\mathbb{Z}[T,N]$ is integral over $\mathbb{Z}$. Since integral dependence is transitive, $\gamma$ is an algebraic integer. This proves the trace and norm integrality criterion for a quadratic extension, including elements already in $F$.

Write $R={\mathcal{O}}_K$. We prove unique factorization of ideals in a number field, rather than assuming this conclusion as a property of a Dedekind domain. Three preliminary facts about $R$ will be useful. Its additive group is a finite-rank free abelian group, so every ideal is finitely generated and the ascending chain condition holds. It is an integrally closed domain, by integral dependence is transitive. Finally, every nonzero prime ideal is a maximal ideal. For this last fact, choose $0\ne a\in P$. Multiplication by $a$ is an injective endomorphism of the finite-rank free abelian group $R$ with nonzero determinant, so $R/(a)$ is finite. Consequently $R/P$ is a finite integral domain and hence a field.

First, every nonzero proper ideal contains a product of nonzero prime ideals. Otherwise choose a maximal counterexample $J$ under inclusion, using the ascending chain condition. It cannot be a prime ideal. Choose $x,y\notin J$ with $xy\in J$. The strictly larger ideals $J+(x)$ and $J+(y)$ each contain a product of nonzero prime ideals, allowing the empty product $R$ when necessary. Multiplying those products gives a product contained ina contradiction.

Next fix a nonzero prime ideal $P$, choose $0\ne a\in P$, and choose a product of nonzero prime idealswith $r$ minimal. Because $(a)\subseteq P$, primality and maximality imply $P_j=P$ for some $j$; reorder so that $P_1=P$. Minimality provides $b\in P_2\cdots P_r$ with $b\notin (a)$, taking the remaining product to be $R$ if $r=1$. ThusDefine the fractional idealIt contains $R$ and $x$. The ideal $P{P}^{-1}$ lies between $P$ and $R$, so it is $P$ or $R$. If it were $P$, then $xP\subseteq P$. Applying the determinant trick to a finite generating set of an ideal for $P$ gives a monic polynomial over $R$ satisfied by $x$. Since $R$ is an integrally closed domain, this would put $x$ in $R$, a contradiction. ThereforeThis proves that each nonzero prime ideal is an invertible fractional ideal, independently of the desired factorization theorem.

Now take a nonzero proper ideal $A$ and a maximal ideal $P$ containing it. The ideal $B=P^{-1}A$ satisfiesThe first inclusion is strict: if $A=B$, the same $x\in P^{-1}\setminus R$ would stabilize $A$, and the determinant trick would again force $x\in R$. Repeating gives a strictly ascending chain until it reaches $R$; the ascending chain condition ensures this happens after finitely many steps. The identities $A=PB$ then express $A$ as a product of nonzero prime ideals.

For uniqueness, suppose $P_1\cdots P_r=Q_1\cdots Q_s$. Since the right-hand side is contained in $P_1$, some $Q_j$ is contained in $P_1$. Both are nonzero maximal ideals, so $Q_j=P_1$. Multiplication by $P_1^{-1}$ cancels this factor. Repeat. A nonempty product of proper prime ideals cannot equal $R$, so both products finish together. This proves uniqueness up to reordering, including repeated factors.

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.